You work for a bottling company. They claim that 16 ounces of beverage goes into every bottle. You have taken a random sample of 100 bottles. You found that the sample mean was 16.1 with a sample standard deviation of 0.4 ounces. You want to determine if the number of ounces in the bottle is different than 16. Perform the appropriate hypothesis test at an alpha level of 0.05
Since we don't know the population standard deviation (only the sample SD) we should use the t-distribution here. Our degrees of freedom will be the sample size, n=100.
First, find the standard error of the sample mean. This will be:
SEM = sample sd / sqrt(n), where n is the sample size.
SEM = 0.4 / sqrt(100) = 0.04.
Next, find the t value associated with our measurement of 16.1. This will be:
(measurement - pop mean) / SEM
t = (16.1-16)/0.04
Now we can go in two different ways (they'll give you the same answer) to test the hypothesis. One, find the probability of seeing a lower t-value than 2.5 for our degrees of freedom:
P(t<2.5, df = 100) = 0.993
Since this is greater than (1 - alpha/2), we reject the null hypothesis. We'd do the same if it was less than alpha/2, i.e. the null hypothesis would be accepted for alpha/2 < P(t) < 1-alpha/2.
Alternatively, you can find the critical t for a 95% confidence interval (100% - alpha) using the inverse. Remember it's two-tailed, so the probability you use is (1-alpha/2), not (1-alpha) since you need the 95% range in the middle so it stretches from 2.5% to 97.5%:
Since the magnitude of the t-stat we found (2.5) is greater than the critical t for the test (1.984), we reject the null hypothesis (note that if we found -2.5, we'd also reject since the magnitude of -2.5 is 2.5).
Mar 24th, 2015
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